Ordered Pairs

i did the same...forgot the negative integers.....

the question should have asked for natural solutions... By the way cant we use this formula in the end..... ${ C }_{ r-1 }^{ n+r-1 }$

Same here hahaha

consider a number $n$ .. we can write it in it's prime factorisation form

$n=p_1^{a_1}p_2^{a_2}p_3^{a_3}...........p_n^{a_n}$

where $p_1,p_2,p_3.....p_n$ are distinct primes.

then, any divisor, $d$ , of $n$ must be of the form

$d=p_1^{b_1}p_2^{b_2}p_3^{b_3}...........p_k^{b_k}$

where $0≤b_i≤a_i$ for $i=1 ,2 ,3 ,......k$

Now, consider the product:

$Q=(1+p_1+p_1^{2}+p_1^{3}+......p_1^{a_1})(1+p_2+p_2^{2}+.........p_2^{a_2}).................(1+p_k+p_k^{2}+..........p_k^{a_k})$

A typical term of this product is $x=p_1^{b_1}......p_k^{b_k}$ for $0≤b_i≤a_i$

Therefore, the number of positive divisors of $n$ is the number of terms in the product $Q$ and since, in each bracket in the product, there are $(a_i+1)$ terms, the total number of terms are:

$(a_1+1)(a_2+1)(a_3+1).........(a_n+1)$

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I think you knew this already, didn't you? :P

nevermind, i brushed up on my latex :P

integers contain positives, negatives and neutral